# C4 Parametrics/Integration Question

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Thread starter 5 years ago
#1
Not entirely sure how to approach this question

Attachment 429025429027

However this is my attempt so far:

My initial idea was to multiply my answer by 4 to find the total area
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5 years ago
#2
(Original post by creativebuzz)
Not entirely sure how to approach this question

Attachment 429025429027

However this is my attempt so far:

My initial idea was to multiply my answer by 4 to find the total area
0
5 years ago
#3
Find the area of the whole shape... then take it away from the square?
damn what paper is this? I have C4 tomorrow and this looks quite hard...
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Thread starter 5 years ago
#4
(Original post by SeanFM)
I suppose finding the area of the square and taking that away could help, but you'd be left with the bits to the left and right of the square, which are in the circle but aren't in the square or the shaded region.

Your idea about multiplying 4 does seem right though. I'm not 100% sure, but the answer may involve finding two integrals (focus on one quadrant and multiply by 4 at the end) and taking one away from the other.
Yeah that's exactly what I was thinking, but I'm still not getting the correct answer (i've plugged it into my graphic calculator)! Perhaps my limits aren't right or?
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5 years ago
#5
(Original post by creativebuzz)
Not entirely sure how to approach this question

Attachment 429025429027

However this is my attempt so far:

My initial idea was to multiply my answer by 4 to find the total area
My thinking is you can turn the whole thing on it's side (treat y as x and vice versa) and do parametric integration from there. The problem is that it's not a function, and I can't remember how that works with regard to integrating for the area (whether it'll end up as 0 or not). Hopefully it gives you the magnitude of the area, in which case this should work, else I can't really think what the answer is!
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Thread starter 5 years ago
#6
(Original post by Duskstar)
My thinking is you can turn the whole thing on it's side (treat y as x and vice versa) and do parametric integration from there. The problem is that it's not a function, and I can't remember how that works with regard to integrating for the area (whether it'll end up as 0 or not). Hopefully it gives you the magnitude of the area, in which case this should work, else I can't really think what the answer is!
As for treating y as x, you can see that I've done that in my working out! But are my limits definitely correct because I'm not entirely sure..
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5 years ago
#7
(Original post by creativebuzz)
As for treating y as x, you can see that I've done that in my working out! But are my limits definitely correct because I'm not entirely sure..
I'm bad at reading other people's workings sorry :P

You've got the limits wrong - when you integrate parametrically you need to redo the limits, because the original limits would be and those are in terms of x, so you need to redefine them in terms of and then it should all work out.

I just did it and got 1.28, I don't know if that's right or not? I can post my working if it is and you want me to ~
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5 years ago
#8
(Original post by creativebuzz)
As for treating y as x, you can see that I've done that in my working out! But are my limits definitely correct because I'm not entirely sure..
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Thread starter 5 years ago
#9
(Original post by Duskstar)
I'm bad at reading other people's workings sorry :P

You've got the limits wrong - when you integrate parametrically you need to redo the limits, because the original limits would be and those are in terms of x, so you need to redefine them in terms of and then it should all work out.

I just did it and got 1.28, I don't know if that's right or not? I can post my working if it is and you want me to ~
The answer is 2.45673...
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5 years ago
#10
(Original post by creativebuzz)
The answer is 2.45673...
Oh.

I'll type up the important parts of my working quickly:

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Thread starter 5 years ago
#11
(Original post by SeanFM)
I can't quite see what your limits are, but they may be right. It's probably what you're integrating and why that's getting difficult.

If we look at the top right quadrant, and integrate between the bottom right point of the shaded area and the top left of that shaded area in that quadrant only, then that should give us the area between (1,root(3)/2), (0,2) and the x axis. If you take what you've just calculated from the integral between (1,0) and (0,2), then you get the area inside the curve that isn't shaded, which leaves you with one more step to find the area of the shaded area given that you have certain integrals and the area of the non-shaded region.
Sean, we strangely very close!

Basically I just shoved 2cosxcosx into my graphic calculator with the limits pi/6 and pi/2 and then multiplied that by 4 which gives me the correct answer however:

a) the answer is negative, which is confusing because I swear the correct order or limits is pi/6 and pi/2, no?

b) why don't we have to subtract the clear squares underneath the curves?

We are on the right lines though!
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5 years ago
#12
(Original post by creativebuzz)
Sean, we strangely very close!

Basically I just shoved 2cosxcosx into my graphic calculator with the limits pi/6 and pi/2 and then multiplied that by 4 which gives me the correct answer however:

a) the answer is negative, which is confusing because I swear the correct order or limits is pi/6 and pi/2, no?

b) why don't we have to subtract the clear squares underneath the curves?

We are on the right lines though!
Oops, what I suggested above was complete rubbish.

I'm not 100% sure about either of those questions so I'll invite anyone else to answer it as I don't want to mislead you.
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5 years ago
#13
I believe this will give you the correct answer:

Try that.
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Thread starter 5 years ago
#14
(Original post by lizard54142)
I believe this will give you the correct answer:

Try that.
Yeah I did that earlier (one sec, lemme copy and paste)

Basically I just shoved 2cosxcosx into my graphic calculator with the limits pi/6 and pi/2 and then multiplied that by 4 which gives me the correct answer however:

a) the answer is negative, which is confusing because I swear the correct order or limits is pi/6 and pi/2, no?

b) why don't we have to subtract the clear squares underneath the curves?
0
5 years ago
#15
(Original post by creativebuzz)
Yeah I did that earlier (one sec, lemme copy and paste)

Basically I just shoved 2cosxcosx into my graphic calculator with the limits pi/6 and pi/2 and then multiplied that by 4 which gives me the correct answer however:

a) the answer is negative, which is confusing because I swear the correct order or limits is pi/6 and pi/2, no?

b) why don't we have to subtract the clear squares underneath the curves?
You have your limits the wrong way round then, mine are the opposite. So the integral I posted must be the correct one
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Thread starter 5 years ago
#16
(Original post by lizard54142)
You have your limits the wrong way round then, mine are the opposite. So the integral I posted must be the correct one
Why is it that way around, surely it should be pi/6 (indicates y=1) and then pi/2 (indicates x= 0)
0
5 years ago
#17
(Original post by lizard54142)
I believe this will give you the correct answer:

Try that.
Do you have any idea where I went wrong? thanks
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5 years ago
#18
(Original post by creativebuzz)
Why is it that way around, surely it should be pi/6 (indicates y=1) and then pi/2 (indicates x= 0)
It's the other way round... start at x=0 and move right along the x axis.
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5 years ago
#19
(Original post by Duskstar)
Do you have any idea where I went wrong? thanks
I'm not to sure what you're trying to do? You seem to have an upper limit of a parameter value and a lower limit of a y value, which won't work
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5 years ago
#20
(Original post by lizard54142)
I'm not to sure what you're trying to do? You seem to have an upper limit of a parameter value and a lower limit of a y value, which won't work
Ah okay I was trying to integrate it as if the x axis was the y axis and vice versa, but I think I got muddled up with the integral's parameters (which should both be parameter values I think). I'll have a look again in the morning, thanks ~
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